Almost sure convergence and Borell - Cantelli Lemma 2
$begingroup$
Suppose we have the following random variable:
$X_n = n$ with probability $frac{1}{n}$ and $0$ with probability $1-frac{1}{n}$. We we can define this variable on the probability space $([0,1], mathcal{B}[0,1], lambda)$ where $lambda$ is the Lebesgue measure. Also, we have independence for all $ngeq1$.
It is said that this $X_n$ converges to $0$ almost surely (several sources). However, once we check this by the Borel - Cantelli Lemma 2, we get that $sum_{n=1}^infty P(X_n=n)=infty$. Given that the events are independent, we know that $X_n=n$ infinitely often.
What could be the reason I am receiving this contradiction?
convergence borel-cantelli-lemmas
asked Dec 17 '18 at 10:24
OviOvi
63
$endgroup$
add a comment |
$begingroup$
Suppose we have the following random variable:
$X_n = n$ with probability $frac{1}{n}$ and $0$ with probability $1-frac{1}{n}$. We we can define this variable on the probability space $([0,1], mathcal{B}[0,1], lambda)$ where $lambda$ is the Lebesgue measure. Also, we have independence for all $ngeq1$.
It is said that this $X_n$ converges to $0$ almost surely (several sources). However, once we check this by the Borel - Cantelli Lemma 2, we get that $sum_{n=1}^infty P(X_n=n)=infty$. Given that the events are independent, we know that $X_n=n$ infinitely often.
What could be the reason I am receiving this contradiction?
convergence borel-cantelli-lemmas
asked Dec 17 '18 at 10:24
OviOvi
63
$endgroup$
$begingroup$
The source is correct, and never asserts that $(X_n)$ is independent (a hypothesis which you seem to have invented by yourself and which does not hold in the setting of the notes you are referring to).
$endgroup$
– Did
Dec 17 '18 at 10:41
$begingroup$
@Did I used the independence, because they do claim it in the same note later. It may have been a typo in the note. However, could you elaborate why the independence does not hold in this setting?
$endgroup$
– Ovi
Dec 17 '18 at 10:52
$begingroup$
Because they explicitely build $(X_n)$ as $X_n=mathbf 1_{(0,1/n)}$ on $[0,1]$ endowed with its Borel sigma-field and the Lebesgue measure, and these are not independent. "they do claim it in the same note later" Sure they do, but for completely different sequences $(X_n)$. No typo here.
$endgroup$
– Did
Dec 17 '18 at 11:44
$begingroup$
@Did Thank you, I completely got it now!
$endgroup$
– Ovi
Dec 17 '18 at 11:54
add a comment |
$begingroup$
Suppose we have the following random variable:
$X_n = n$ with probability $frac{1}{n}$ and $0$ with probability $1-frac{1}{n}$. We we can define this variable on the probability space $([0,1], mathcal{B}[0,1], lambda)$ where $lambda$ is the Lebesgue measure. Also, we have independence for all $ngeq1$.
It is said that this $X_n$ converges to $0$ almost surely (several sources). However, once we check this by the Borel - Cantelli Lemma 2, we get that $sum_{n=1}^infty P(X_n=n)=infty$. Given that the events are independent, we know that $X_n=n$ infinitely often.
What could be the reason I am receiving this contradiction?
convergence borel-cantelli-lemmas
asked Dec 17 '18 at 10:24
OviOvi
63
$endgroup$
Suppose we have the following random variable:
$X_n = n$ with probability $frac{1}{n}$ and $0$ with probability $1-frac{1}{n}$. We we can define this variable on the probability space $([0,1], mathcal{B}[0,1], lambda)$ where $lambda$ is the Lebesgue measure. Also, we have independence for all $ngeq1$.
It is said that this $X_n$ converges to $0$ almost surely (several sources). However, once we check this by the Borel - Cantelli Lemma 2, we get that $sum_{n=1}^infty P(X_n=n)=infty$. Given that the events are independent, we know that $X_n=n$ infinitely often.
What could be the reason I am receiving this contradiction?
convergence borel-cantelli-lemmas
convergence borel-cantelli-lemmas
asked Dec 17 '18 at 10:24
OviOvi
63
asked Dec 17 '18 at 10:24
OviOvi
63
asked Dec 17 '18 at 10:24
OviOvi
63
asked Dec 17 '18 at 10:24
OviOvi
63
asked Dec 17 '18 at 10:24
OviOvi
63
63
$begingroup$
The source is correct, and never asserts that $(X_n)$ is independent (a hypothesis which you seem to have invented by yourself and which does not hold in the setting of the notes you are referring to).
$endgroup$
– Did
Dec 17 '18 at 10:41
$begingroup$
@Did I used the independence, because they do claim it in the same note later. It may have been a typo in the note. However, could you elaborate why the independence does not hold in this setting?
$endgroup$
– Ovi
Dec 17 '18 at 10:52
$begingroup$
Because they explicitely build $(X_n)$ as $X_n=mathbf 1_{(0,1/n)}$ on $[0,1]$ endowed with its Borel sigma-field and the Lebesgue measure, and these are not independent. "they do claim it in the same note later" Sure they do, but for completely different sequences $(X_n)$. No typo here.
$endgroup$
– Did
Dec 17 '18 at 11:44
$begingroup$
@Did Thank you, I completely got it now!
$endgroup$
– Ovi
Dec 17 '18 at 11:54
add a comment |
$begingroup$
The source is correct, and never asserts that $(X_n)$ is independent (a hypothesis which you seem to have invented by yourself and which does not hold in the setting of the notes you are referring to).
$endgroup$
– Did
Dec 17 '18 at 10:41
$begingroup$
@Did I used the independence, because they do claim it in the same note later. It may have been a typo in the note. However, could you elaborate why the independence does not hold in this setting?
$endgroup$
– Ovi
Dec 17 '18 at 10:52
$begingroup$
Because they explicitely build $(X_n)$ as $X_n=mathbf 1_{(0,1/n)}$ on $[0,1]$ endowed with its Borel sigma-field and the Lebesgue measure, and these are not independent. "they do claim it in the same note later" Sure they do, but for completely different sequences $(X_n)$. No typo here.
$endgroup$
– Did
Dec 17 '18 at 11:44
$begingroup$
@Did Thank you, I completely got it now!
$endgroup$
– Ovi
Dec 17 '18 at 11:54
$begingroup$
The source is correct, and never asserts that $(X_n)$ is independent (a hypothesis which you seem to have invented by yourself and which does not hold in the setting of the notes you are referring to).
$endgroup$
– Did
Dec 17 '18 at 10:41
$begingroup$
The source is correct, and never asserts that $(X_n)$ is independent (a hypothesis which you seem to have invented by yourself and which does not hold in the setting of the notes you are referring to).
$endgroup$
– Did
Dec 17 '18 at 10:41
$begingroup$
@Did I used the independence, because they do claim it in the same note later. It may have been a typo in the note. However, could you elaborate why the independence does not hold in this setting?
$endgroup$
– Ovi
Dec 17 '18 at 10:52
$begingroup$
@Did I used the independence, because they do claim it in the same note later. It may have been a typo in the note. However, could you elaborate why the independence does not hold in this setting?
$endgroup$
– Ovi
Dec 17 '18 at 10:52
$begingroup$
Because they explicitely build $(X_n)$ as $X_n=mathbf 1_{(0,1/n)}$ on $[0,1]$ endowed with its Borel sigma-field and the Lebesgue measure, and these are not independent. "they do claim it in the same note later" Sure they do, but for completely different sequences $(X_n)$. No typo here.
$endgroup$
– Did
Dec 17 '18 at 11:44
$begingroup$
Because they explicitely build $(X_n)$ as $X_n=mathbf 1_{(0,1/n)}$ on $[0,1]$ endowed with its Borel sigma-field and the Lebesgue measure, and these are not independent. "they do claim it in the same note later" Sure they do, but for completely different sequences $(X_n)$. No typo here.
$endgroup$
– Did
Dec 17 '18 at 11:44
$begingroup$
@Did Thank you, I completely got it now!
$endgroup$
– Ovi
Dec 17 '18 at 11:54
$begingroup$
@Did Thank you, I completely got it now!
$endgroup$
– Ovi
Dec 17 '18 at 11:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is an answer to the question as posted here. The random variables are not defined explicitly in this question. It is just mentioned that they are defined on $(0,1)$ with Lebesgue measure. Explicit definition was given in the comments after I posted this answer.
I don't know what those sources are but we can only conclude that $X_n to 0$ in probability. It need not converge almost surely.
answered Dec 17 '18 at 10:34
Kavi Rama MurthyKavi Rama Murthy
68k53068
$endgroup$
$begingroup$
Thanks for the comment! Here is the link: nptel.ac.in/courses/108106083/lecture28_Convergence.pdf). Page 3, Example. I also think that, clearly, we can only talk about convergence in probability.
$endgroup$
– Ovi
Dec 17 '18 at 10:39
$begingroup$
See comment on main.
$endgroup$
– Did
Dec 17 '18 at 10:41
add a comment |
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1 Answer
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1 Answer
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$begingroup$
This is an answer to the question as posted here. The random variables are not defined explicitly in this question. It is just mentioned that they are defined on $(0,1)$ with Lebesgue measure. Explicit definition was given in the comments after I posted this answer.
I don't know what those sources are but we can only conclude that $X_n to 0$ in probability. It need not converge almost surely.
answered Dec 17 '18 at 10:34
Kavi Rama MurthyKavi Rama Murthy
68k53068
$endgroup$
$begingroup$
Thanks for the comment! Here is the link: nptel.ac.in/courses/108106083/lecture28_Convergence.pdf). Page 3, Example. I also think that, clearly, we can only talk about convergence in probability.
$endgroup$
– Ovi
Dec 17 '18 at 10:39
$begingroup$
See comment on main.
$endgroup$
– Did
Dec 17 '18 at 10:41
add a comment |
$begingroup$
This is an answer to the question as posted here. The random variables are not defined explicitly in this question. It is just mentioned that they are defined on $(0,1)$ with Lebesgue measure. Explicit definition was given in the comments after I posted this answer.
I don't know what those sources are but we can only conclude that $X_n to 0$ in probability. It need not converge almost surely.
answered Dec 17 '18 at 10:34
Kavi Rama MurthyKavi Rama Murthy
68k53068
$endgroup$
$begingroup$
Thanks for the comment! Here is the link: nptel.ac.in/courses/108106083/lecture28_Convergence.pdf). Page 3, Example. I also think that, clearly, we can only talk about convergence in probability.
$endgroup$
– Ovi
Dec 17 '18 at 10:39
$begingroup$
See comment on main.
$endgroup$
– Did
Dec 17 '18 at 10:41
add a comment |
$begingroup$
This is an answer to the question as posted here. The random variables are not defined explicitly in this question. It is just mentioned that they are defined on $(0,1)$ with Lebesgue measure. Explicit definition was given in the comments after I posted this answer.
I don't know what those sources are but we can only conclude that $X_n to 0$ in probability. It need not converge almost surely.
answered Dec 17 '18 at 10:34
Kavi Rama MurthyKavi Rama Murthy
68k53068
$endgroup$
This is an answer to the question as posted here. The random variables are not defined explicitly in this question. It is just mentioned that they are defined on $(0,1)$ with Lebesgue measure. Explicit definition was given in the comments after I posted this answer.
I don't know what those sources are but we can only conclude that $X_n to 0$ in probability. It need not converge almost surely.
answered Dec 17 '18 at 10:34
Kavi Rama MurthyKavi Rama Murthy
68k53068
edited Dec 17 '18 at 11:48
answered Dec 17 '18 at 10:34
Kavi Rama MurthyKavi Rama Murthy
68k53068
answered Dec 17 '18 at 10:34
Kavi Rama MurthyKavi Rama Murthy
68k53068
answered Dec 17 '18 at 10:34
Kavi Rama MurthyKavi Rama Murthy
68k53068
68k53068
$begingroup$
Thanks for the comment! Here is the link: nptel.ac.in/courses/108106083/lecture28_Convergence.pdf). Page 3, Example. I also think that, clearly, we can only talk about convergence in probability.
$endgroup$
– Ovi
Dec 17 '18 at 10:39
$begingroup$
See comment on main.
$endgroup$
– Did
Dec 17 '18 at 10:41
add a comment |
$begingroup$
Thanks for the comment! Here is the link: nptel.ac.in/courses/108106083/lecture28_Convergence.pdf). Page 3, Example. I also think that, clearly, we can only talk about convergence in probability.
$endgroup$
– Ovi
Dec 17 '18 at 10:39
$begingroup$
See comment on main.
$endgroup$
– Did
Dec 17 '18 at 10:41
$begingroup$
Thanks for the comment! Here is the link: nptel.ac.in/courses/108106083/lecture28_Convergence.pdf). Page 3, Example. I also think that, clearly, we can only talk about convergence in probability.
$endgroup$
– Ovi
Dec 17 '18 at 10:39
$begingroup$
Thanks for the comment! Here is the link: nptel.ac.in/courses/108106083/lecture28_Convergence.pdf). Page 3, Example. I also think that, clearly, we can only talk about convergence in probability.
$endgroup$
– Ovi
Dec 17 '18 at 10:39
$begingroup$
See comment on main.
$endgroup$
– Did
Dec 17 '18 at 10:41
$begingroup$
See comment on main.
$endgroup$
– Did
Dec 17 '18 at 10:41
add a comment |
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$begingroup$
The source is correct, and never asserts that $(X_n)$ is independent (a hypothesis which you seem to have invented by yourself and which does not hold in the setting of the notes you are referring to).
$endgroup$
– Did
Dec 17 '18 at 10:41
$begingroup$
@Did I used the independence, because they do claim it in the same note later. It may have been a typo in the note. However, could you elaborate why the independence does not hold in this setting?
$endgroup$
– Ovi
Dec 17 '18 at 10:52
$begingroup$
Because they explicitely build $(X_n)$ as $X_n=mathbf 1_{(0,1/n)}$ on $[0,1]$ endowed with its Borel sigma-field and the Lebesgue measure, and these are not independent. "they do claim it in the same note later" Sure they do, but for completely different sequences $(X_n)$. No typo here.
$endgroup$
– Did
Dec 17 '18 at 11:44
$begingroup$
@Did Thank you, I completely got it now!
$endgroup$
– Ovi
Dec 17 '18 at 11:54